Sum (sin(Pi*n)/(-1+n^2) , n=1infinity). n=1,2,3

[GerardEncina]

Hi everybody!

Solving a PDE for the damped wave equation I get with a solutions that part of it have the expression:

Sum (sin(Pi*n)/(-1+n^2) , n=1..infinity). n=1,2,3...

When calculating it with Maple I get this is equal to -1/2*Pi.

Can someone of you explain me why?

Thanks a lot!

 

[LCKurtz]

Hi everybody!

Solving a PDE for the damped wave equation I get with a solutions that part of it have the expression:

Sum (sin(Pi*n)/(-1+n^2) , n=1..infinity). n=1,2,3...

When calculating it with Maple I get this is equal to -1/2*Pi.

Can someone of you explain me why?

Thanks a lot!

Did you get that expression from a Fourier Series expansion? Sometimes these sums arise by simply evaluating the FS at some point.

 

[GerardEncina]

Yes, I'm trying to solve the PDE for the damped wave equations defined like:

u_tt+u_t-u_xx=0,

With Initial Conditions: u(x,0)=0 , u_t (x,0)=sin⁡(x)
And Boundary Conditions u(0,t)=u(Pi,t)=0

Finally I end with the solution u(x,t)=∑(n=1 to ∞) [exp^(-t/2)*(-4sin⁡(Pi*n))*sin⁡(sqrt(4n^2-1)*t/2)*sin(nx) / (Pi*sqrt(4n^2-1)*(-1+n^2))]

From here, there is the expression sin(Pi*n)/(-1+n^2), that for n=1 is 0. But solving it with Maple it gives u(x,t)=2/(3*sqrt(3))*exp^(-t/2)*sin(sqrt(3)*t/2)*sin(x).

I've also found that limit [when n-->1] of sin(Pi*n)/(-1+n^2) = -Pi/2; that plugged on the series give what Maple says.

But I'm still not sure if it is correct or not

Thanks!

 

[Mute]

Yes, I'm trying to solve the PDE for the damped wave equations defined like:

u_tt+u_t-u_xx=0,

With Initial Conditions: u(x,0)=0 , u_t (x,0)=sin⁡(x)
And Boundary Conditions u(0,t)=u(Pi,t)=0

Finally I end with the solution u(x,t)=∑(n=1 to ∞) [exp^(-t/2)*(-4sin⁡(Pi*n))*sin⁡(sqrt(4n^2-1)*t/2)*sin(nx) / (Pi*sqrt(4n^2-1)*(-1+n^2))]

From here, there is the expression sin(Pi*n)/(-1+n^2), that for n=1 is 0. But solving it with Maple it gives u(x,t)=2/(3*sqrt(3))*exp^(-t/2)*sin(sqrt(3)*t/2)*sin(x).

I've also found that limit [when n-->1] of sin(Pi*n)/(-1+n^2) = -Pi/2; that plugged on the series give what Maple says.

But I'm still not sure if it is correct or not

Thanks!

\sin(n\pi) is zero for any integer n. So, all of your terms except for the first one is zero. The first is treated as non-zero because the denominator vanishes at n = 1. So, you use L'Hopital's rule to evaluate it instead.